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Algebra Difficulty 6.1 National olympiad Prove it

Example 7 Let 0<t1⩽t2⩽⋯⩽tn<10<t_{1} \leqslant t_{2} \leqslant \cdots \leqslant t_{n}<1. Prove:
(1−tn)2[t1(1−t12)2+t22(1−t23)2+⋯+tnn(1−tnn+1)2]<1\left(1-t_{n}\right)^{2}\left[\frac{t_{1}}{\left(1-t_{1}^{2}\right)^{2}}+\frac{t_{2}^{2}}{\left(1-t_{2}^{3}\right)^{2}}+\cdots+\frac{t_{n}^{n}}{\left(1-t_{n}^{n+1}\right)^{2}}\right]<1.
(IMO - 28 Shortlist)

Solution

Given by the problem, we know that (1−tn)2(1−tk)2⩽1,k=1,2,⋯ ,n\frac{\left(1-t_{n}\right)^{2}}{\left(1-t_{k}\right)^{2}} \leqslant 1, k=1,2, \cdots, n, so
t1(1−tn)2(1−t12)2⩽t1(1+t1)2<t11+t1=1−11+t1,t22(1−tn)2(1−t23)2⩽t22(1+t2+t22)2<11+t2−11+t1+t22⩽11+t1−11+t2+t22,tkk(1−tn)2(1−tkk+1)2⩽1(1+tk+⋯+tkk)2<11+tk+⋯+tkk−1−11+tk+⋯+tkk⩽11+tk−1+tk−12+⋯+tk−1k−1−11+tk+⋯+tkk, where k=3,4,⋯ \begin{aligned} \frac{t_{1}\left(1-t_{n}\right)^{2}}{\left(1-t_{1}^{2}\right)^{2}} & \leqslant \frac{t_{1}}{\left(1+t_{1}\right)^{2}}<\frac{t_{1}}{1+t_{1}}=1-\frac{1}{1+t_{1}}, \\ \frac{t_{2}^{2}\left(1-t_{n}\right)^{2}}{\left(1-t_{2}^{3}\right)^{2}} & \leqslant \frac{t_{2}^{2}}{\left(1+t_{2}+t_{2}^{2}\right)^{2}}<\frac{1}{1+t_{2}}-\frac{1}{1+t_{1}+t_{2}^{2}} \leqslant \frac{1}{1+t_{1}}-\frac{1}{1+t_{2}+t_{2}^{2}}, \\ \frac{t_{k}^{k}\left(1-t_{n}\right)^{2}}{\left(1-t_{k}^{k+1}\right)^{2}} & \leqslant \frac{1}{\left(1+t_{k}+\cdots+t_{k}^{k}\right)^{2}}<\frac{1}{1+t_{k}+\cdots+t_{k}^{k-1}}-\frac{1}{1+t_{k}+\cdots+t_{k}^{k}} \\ & \leqslant \frac{1}{1+t_{k-1}+t_{k-1}^{2}+\cdots+t_{k-1}^{k-1}}-\frac{1}{1+t_{k}+\cdots+t_{k}^{k}} \text {, where } k=3,4, \cdots \end{aligned}

Adding both sides of the above inequalities, we obtain the desired inequality.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.